Question

Help me prove this!Will medal!
Prove using the definition of the limit:
lim as x approaches -2 of (x^2 + 3x - 1) = -3

Answer 1

\[\lim_{x \rightarrow -2} (x^2 + 3x -1) = -3\]

Answer 2

this is a continuous function on the point in question. i.e. plug it in

Answer 3

Yeah, I know the limit's value but I need to prove it using the definition of the limit.

Definition of limit: Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a. Then we say that,
\[\lim_{x \rightarrow a}f(x) = L\]
if for every number epsilon > 0, there is some number delta > 0, such that

\[\left| f(x) - L \right| < \epsilon , whenever 0 < \left| x- a \right| < \delta\]

Answer 4

Ew. Haven't done these in a while.

\(\large |(x^2+3x-1) - (-3)| < \epsilon\) wherever \(\large 0 < \left| x- (-2) \right| < \delta\)

simplify and factor

\(\large |x^2+3x+2| < \epsilon\) wherever \(\large 0 < \left| x+2 \right| < \delta\)


\(\large |x+2| |x+1| < \epsilon\) wherever \(\large 0 < \left| x+2 \right| < \delta\)

Answer 5

I know about that, but you need to get rid of x+2 in order to prove it.That's where I'm having trouble.But thanks.

Answer 6

I meant x+1.

Answer 7

its continuous

you "prove it" by plugging it in

Answer 8

No, in this case "proving" means using the definition of the limit, to riguriously state that.
Simply saying that that's true doesn't prove it.

Answer 9

Example 3: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx

Answer 10

you can plug in

\(x = (- 2 + h)_{h \to 0} \) if you like with \(0 < |h| << 0\)


but not here

Answer 11

Lol Paul's online math notes are exactly where I am learning from. I don't quite get their explanation. Why do we set x+5 smaller than some number K?

Answer 12

\[\large k |x+1| < \epsilon\]so that you can do it, like all the other examples, and write it as \[\large |x+1| < \frac{ \epsilon }{ k } \]

Answer 13

Yeah, but here you substituting K for the x+5, isn't K bigger than x+5?

Answer 14

Oops, should be x+1

Answer 15

Oh, I get it now. I have another question, why do you pick the minimum of the 1, epsilon/10?

Answer 16

We're using \(\large |x+1|<k\) so that
\[\large |x+2||x+1|<k|x+2|\]

Answer 17

I think Paul's notes explain that well.

Answer 18

What does guaranteeing that delta is smaller than or equal to epsilon/10 do?
Everything doesn't quite piece together in my head.

Answer 19

So like